Question

A 10 m long wire of resistance 20 $ \Omega $ is connected in series with a battery of emf 3V and a resistance of 10 $ \Omega $ . The potential gradient along the wire in V/m is

• A. 0.02
• B. 0.1
• C. 0.2
• D. 1.2

Answer 1

Answer: (C)

As resistances are in series $ {{R}_{total}}={{R}_{1}}+{{R}_{2}} $ $ =20+10=30\Omega $ $ i=\frac{4}{{{R}_{\text{total}}}}=\frac{3}{30}=\frac{1}{10}\text{A} $ So $ {{V}_{\text{wire}}}=\,\,i{{R}_{\text{wire}}} $ $ \Rightarrow $ $ {{V}_{\text{wire}}}=\frac{1}{10}\times 20=2V $ Hence, potential gradient is $ \frac{{{V}_{\text{wire}}}}{l}=\frac{2}{10}=0.2\,\,\text{V/m} $